“Chapter4”

Access the required packages

library(MASS)
library(GGally)
library(ggplot2)
library(tidyr)
library(corrplot)

## corrplot 0.84 loaded

Loaded the Boston dataset

data("Boston")

The Boston dataset shows housing values in suburbs of Boston with variables that are related to the value of housing such as per capita crime rate by town, the proportion of residential land zoned for lots over 25,000 sq.ft, the proportion of non-retail business acres per town, proportion of owner-occupied units built prior to 1940, full-value property-tax rate per $10,000, pupil-teacher ratio by town, median value of owner-occupied homes in $1000s.

Explored the Boston dataset

str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

dim(Boston)

## [1] 506  14

summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

The Boston data frame has 506 rows and 14 columns.

Graphical Overview of the data.

calculate the correlation matrix and round it

cor_matrix<-cor(Boston) 

Visualized the correlation matrix

The function corrplot function was used to can be used to visualize the correlations.

corrplot(cor_matrix, method="circle")

cor_matrix<-cor(Boston) %>% round(digits = 2)
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)

Positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the circle are proportional to the correlation coefficients. In the right side of the correlogram, the legend color shows the correlation coefficients and the corresponding colors. The corrplot shows the correlations between variables of the Boston dataset. From the plot see that that rad (index of accessibility to radial highways) and tax (full-value property-tax rate per $10000) have the strongest positive correlation between the variables. While dis (weighted mean of distances to five Boston employment centres) and nox (nitrogen oxides concentration), age (proportion of owner-occupied units built prior to 1940) and indux (proportion of non-retail business acres per town), medv (meidan value of owner-occupied homes in $1000) and lstat (lower status of the population) have very strong negative correlation. # Printed the correlation matrix

cor_matrix

##          crim    zn indus  chas   nox    rm   age   dis   rad   tax
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47
##         ptratio black lstat  medv
## crim       0.29 -0.39  0.46 -0.39
## zn        -0.39  0.18 -0.41  0.36
## indus      0.38 -0.36  0.60 -0.48
## chas      -0.12  0.05 -0.05  0.18
## nox        0.19 -0.38  0.59 -0.43
## rm        -0.36  0.13 -0.61  0.70
## age        0.26 -0.27  0.60 -0.38
## dis       -0.23  0.29 -0.50  0.25
## rad        0.46 -0.44  0.49 -0.38
## tax        0.46 -0.44  0.54 -0.47
## ptratio    1.00 -0.18  0.37 -0.51
## black     -0.18  1.00 -0.37  0.33
## lstat      0.37 -0.37  1.00 -0.74
## medv      -0.51  0.33 -0.74  1.00

Summary of cor_matrix variables.

# center and standardize variables
boston_scaled <- scale(Boston)

# summaries of the scaled variables
summary(boston_scaled)

##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

After scaling the variables, the means of each variables have become very close to zero. Many of the minimum and maximum values for each variable have decreased.

Next we place the variables into quantiles, to get high, med_low, med_high, and low rates of crime into their own categories

# class of the boston_scaled object
class(boston_scaled)

## [1] "matrix"

# change the object to data frame
boston_scaled<-as.data.frame(boston_scaled)

# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)

bins

##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610

First, we use the quantiles; low, med_low, med_high, and high as the break points and create a categorical variable of the crime rate in the Boston dataset.

# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))

# look at the table of the new factor crime
table(crime)

## crime
##      low  med_low med_high     high 
##      127      126      126      127

# summary of the scaled crime rate
summary(crime)

##      low  med_low med_high     high 
##      127      126      126      127

Then the old crime rate variable is removed and then the dataset is divided into train and test so that 80% of the data belongs to the train set.

# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# boston_scaled is available

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

Linear discriminant analysis

Linear Discriminant analysis is a classification method that finds the linear combination of the variables that separate the target variable classes.

Here I fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Drew the LDA (bi)plot.

# linear discriminant analysis
lda.fit <- lda(crime~., data = train)

# print the lda.fit object
lda.fit

## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2524752 0.2500000 0.2475248 0.2500000 
## 
## Group means:
##                  zn      indus        chas        nox          rm
## low       0.9878174 -0.9220520 -0.15653200 -0.8868460  0.51316950
## med_low  -0.0752376 -0.3144696 -0.03844192 -0.5680453 -0.15472665
## med_high -0.3826200  0.1881567  0.20012296  0.3549859  0.06096579
## high     -0.4872402  1.0171306 -0.07742312  1.0518773 -0.43025943
##                 age        dis        rad        tax      ptratio
## low      -0.8790938  0.8327825 -0.6812431 -0.7437234 -0.446800297
## med_low  -0.3311831  0.3805371 -0.5315812 -0.4606297 -0.006443321
## med_high  0.3981368 -0.3516731 -0.3904107 -0.2934496 -0.270461123
## high      0.8299353 -0.8461561  1.6379981  1.5139626  0.780625165
##                black       lstat        medv
## low       0.38483134 -0.79464823  0.58954379
## med_low   0.32005236 -0.08539289 -0.04425999
## med_high  0.07279017  0.03404771  0.13343262
## high     -0.72667022  0.88353979 -0.70245254
## 
## Coefficients of linear discriminants:
##                 LD1          LD2         LD3
## zn       0.07126159  0.851038594 -0.88420600
## indus    0.03129580 -0.271096511  0.15041977
## chas    -0.05577376 -0.139950376  0.09749537
## nox      0.40119425 -0.639081855 -1.27817456
## rm      -0.07397862 -0.054079981 -0.12944887
## age      0.29830317 -0.305633691 -0.21936546
## dis     -0.01814237 -0.417839478  0.26408882
## rad      2.92407923  1.047827512 -0.21174919
## tax      0.06258293 -0.210981352  0.78484498
## ptratio  0.09549014  0.098676374 -0.19309405
## black   -0.12618047  0.007673174  0.13779207
## lstat    0.24007739 -0.248433380  0.41761752
## medv     0.19829640 -0.364569576 -0.25719686
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9435 0.0416 0.0149

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

Next I saved the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then I predicted the classes with the LDA model on the test data. I Cross tabulated the results with the crime categories from the test set. Commented on the results.

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)

# lda.fit, correct_classes and test are available

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)

##           predicted
## correct    low med_low med_high high
##   low       13      10        2    0
##   med_low    5      15        5    0
##   med_high   0       8       18    0
##   high       0       0        0   26

# lda.fit, correct_classes and test are available

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class) %>% addmargins

##           predicted
## correct    low med_low med_high high Sum
##   low       13      10        2    0  25
##   med_low    5      15        5    0  25
##   med_high   0       8       18    0  26
##   high       0       0        0   26  26
##   Sum       18      33       25   26 102

# Percentage inaccuracy
table(correct = correct_classes, predicted = lda.pred$class) %>% prop.table() %>% addmargins()

##           predicted
## correct           low    med_low   med_high       high        Sum
##   low      0.12745098 0.09803922 0.01960784 0.00000000 0.24509804
##   med_low  0.04901961 0.14705882 0.04901961 0.00000000 0.24509804
##   med_high 0.00000000 0.07843137 0.17647059 0.00000000 0.25490196
##   high     0.00000000 0.00000000 0.00000000 0.25490196 0.25490196
##   Sum      0.17647059 0.32352941 0.24509804 0.25490196 1.00000000

loss_func <- function(class, prob) {
  n_wrong <- class != prob
  mean(n_wrong)
}
loss_func(class = correct_classes, prob = lda.pred$class)

## [1] 0.2941176

From the cross table, it is shown that most of the predictions are correct and from the total proportion of inaccurately predicted individuals the prediction model is close to 31% incorrect, making the model close to 69% accurate. This is a relatively good model based on these results.

Here I reloaded the Boston dataset and standardize the dataset (we did not do this in the Datacamp exercises, but you should scale the variables to get comparable distances). Calculated the distances between the observations.

# load MASS and Boston
library(MASS)
data('Boston')

#scaled
boston_scaled=scale(Boston)

# euclidean distance matrix
dist_eu <- dist(boston_scaled)

# look at the summary of the distances
summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

# manhattan distance matrix
dist_man <- dist(boston_scaled, method = 'manhattan')

# look at the summary of the distances
summary(dist_man)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2662  8.4832 12.6090 13.5488 17.7568 48.8618

summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Ran the k-means algorithm on the dataset. Investigated what is the optimal number of clusters and reran the algorithm again. Visualized the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results.

# k-means clustering
km <-kmeans(boston_scaled, centers = 3)

# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)

# MASS, ggplot2 and Boston dataset are available
set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

# k-means clustering
km <-kmeans(boston_scaled, centers = 2)

# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)

The graph shows that 2 clusters are the optimal number of clusters. The total of within cluster sum of squares (WCSS) vs. the number of cluster changes is plotted. We can observe the clusters on the data and how they relate in combination for each variable.

Bonus:

Perform k-means on the original Boston data with some reasonable number of clusters (> 2). Remember to standardize the dataset. Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model. Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution).

library(MASS)
data('Boston')

#scaled
boston_scaledx=scale(Boston)
km2 <-kmeans(boston_scaledx, centers = 3)
boston_scaledx <- as.data.frame(boston_scaledx)
#LDA
lda.fit2 <- lda(km2$cluster ~., data=boston_scaledx)

#lda biplot 
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(km2$cluster)

# plot the lda results
plot(lda.fit2, dimen = 2, col=classes, pch= classes)
lda.arrows(lda.fit, myscale = 1)

It appears that rad (index of accessibility to radial highways) is the most influencial linear separators for the clusters, as well as nox (nitrogen oxides concentration) and (zn) proportion of residential land zoned for lots over 25,000 sq.ft. #Super Bonus

#First run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

library(plotly)

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
## 
##     last_plot

## The following object is masked from 'package:MASS':
## 
##     select

## The following object is masked from 'package:stats':
## 
##     filter

## The following object is masked from 'package:graphics':
## 
##     layout

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

The first graphic is not that clear. There are two clusters we can identify which observations belong to which crime classes.

In the second graphic we observe the graphic where there are 4 colors, one color for each cluster; green is low, med_low is orange, blue is med_high, pink is high. The optimal cluster number is 2 clusters.